Signal Processing and Linear Systems-B.P.Lathi copy

# Note t hat yot b eing t he z ero input component jt

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: own below: A2 + 4A + 40 = (A2 + 4A + 4) + 36 = (A + 2)2 + (6)2 = (A + 2 - j6)(A + 2 + j 6) llO 2 T ime-Domain A nalysis o f C ontinuous-Time S ystems where c and (J are arbitrary constants to be determined from the initial conditions yo(O) = 2 a nd yo(O) = 16.78. Differentiation of Eq. (2.12a) yields yo(t) = _ 2ce- 2t cos (6t + (J) 6ce- 2t sin (6t + Ii) (2.12b) S etting t = 0 in Eqs. (2.12a) and (2.12b), and then substituting initial conditions, we o btain 2 =ccosli - 16.78 = - 2ccos Ii - 6csin Ii Solution of these two simultaneous equations in two unknowns ccos ccos (J = 2 (J and csin csin (J = - 3.463 Squaring and t hen adding the two sides of the above equations yields 2 = (2)2 + ( -3.464)2 = 16 ==&gt; c = 4 c Next, dividing (2.13b) by (2.13a); t hat is dividing csin t an (J = - 3 463 2 and (J Therefore by ccos (J yields (2.13a) (2.13b) yields = t an- 1 (-3 2 463) = -1 yo(t) = 4 e- 2t cos (6t Figure B .llc shows the plot of yo(t). o (J (J C omputer E xample C 2.1 Find the r oots of polynomial &gt;-2 - 1) • + 4&gt;- + 40 a =[l 4 40]; r = r oots(a) o r --2.0000 + 6 .0000i - 2.0000 - 6.0000i 0 C omputer E xample C 2.2 For an LTIC system specified by the differential equation (D2 ( b) ( e) y O=dsolve('D2y+4*Dy+3*y=0','y(0)=3','Dy(0)=-7','t') yO = 2 *exp(-3*t)+exp(-t) y O=dsolve('D2y+4*Dy+4*y=O' , 'y(O)=3', ' Dy(O)=-7', ' t') yO = 3 *exp(-2*t)-exp(-2t)*t y O=dsolve('D2y+4*Dy+40*y=O' , 'y(O)=3' , 'Dy(0)=-7', ' t') yO = 3 *exp(-2*t)*eos(6*t)-1/6*exp(-2t)*sin(6*t) 0 6. E xercise E 2.1 F ind t he z ero-input r esponse o f a n LTIC s ystem described by (D+ 5)y(t) = f (t) i f t he i nitial condition is y(O) = 5. Answer: yo(t) = 5 e- 5t t:::: 0 'V 6. E xercise E 2.2 Solve (D2 + 2D) yo(t) = 0 i f Yo(O) = 1 a nd !io(O) = 4. Hint: T he c haracteristic roots a re 0 a nd - 2. Answer: yo(t) = 3 - 2e- 2t t :::: 0 'V S ystem R esponse t o I nternal C onditions: Z ero-Input R esponse 111 Practical Initial Conditions and the meaning of 0 - and 0 + I n E xample 2.1 t he i nitial c onditions YolO) a nd yolO) were supplied. I n p racticl\l p roblems, we m ust d erive such conditions from t he p hysical s ituation. F or instance, i n a n R LC c ircuit, we m ay b e g iven t he c onditions, s uch as i nitial c apacitor v oltages, a nd i nitial i nductor c urrents, e tc. F rom t his i nformation, we need t o d erive yolO), yolO), . .. for t he d esired variable a s d emonstrated i n t he n ext e xample. I n m uch o f o ur d iscussion, t he i nput is a ssumed t o s tart a t t = 0, u nless o therwise mentioned. Hence, t = 0 is t he r eference p oint. T he c onditions i mmediately b efore t = 0 ( just b efore t he i nput is applied) a re t he c onditions a t t = 0 -, a nd t hose i mmediately a fter t = 0 ( just a fter t he i nput is applied) a re t he c onditions a t t = 0 + ( compare t his w ith t he h istorical t ime f rame B .C. a nd A .D.). I n p ractice, we a re likely t o k now t he i nitial c onditions a t t = 0 - rather t han a t t = 0+. T he t wo s ets o f conditions a re g enerally different, a lthough i n s ome cases t hey m ay b e i dentical. W e a re d ealing w ith t he t otal r esponse y (t), w hich c onsists o f two c omponents; t he z ero-input...
View Full Document

Ask a homework question - tutors are online